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6. Given $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$
Then the number of ordered pairs $(a, r)$ that satisfy the condition is
|
6. 62.
From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$
Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.
By the universal formula for two angles.
The probability of returning to the starting point after $n$ steps is
$$
\begin{array}{l}
\left.\frac{5}{3}=\frac{\cos A \cdot \mathrm{c}}{\sin A}-p_{n}\right) \\
\Rightarrow p_{n}=\frac{2}{3}\left(-\frac{1}{2}\right)^{n}+\frac{1}{3} .
\end{array}
$$
Substituting $n=10$ into the above formula yields.
|
62
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$.
|
【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$
This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^{2}}{c}+\frac{b}{c^{2}}=1 \text { becomes } \\
x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}=0 . \\
\text { Hence } a b^{5}+b c^{5}+c a^{5}-3 \\
=\frac{x y^{4}}{z^{5}}+\frac{y z^{4}}{x^{5}}+\frac{z x^{4}}{y^{5}}-3 \\
=\frac{x^{6} y^{9}+y^{6} z^{9}+z^{6} x^{9}-3 x^{5} y^{5} z^{5}}{x^{5} y^{5} z^{5}} \\
=\frac{\left(x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}\right)\left(x^{4} y^{6}+y^{4} z^{6}+z^{4} x^{6}-x^{2} y^{5} z^{3}-y^{2} z^{5} x^{3}-z^{2} x^{5} y^{3}\right)}{x^{5} y^{5}} \\
=0 \\
\Rightarrow a b^{5}+b c^{5}+c a^{5}=3 .
\end{array}
$$
Solution 2 From $a b c=-1 \Rightarrow b=-\frac{1}{a c}$.
Substitute into $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$, we get $a^{3} c^{2}=a c^{3}+1$.
$$
\begin{array}{l}
\text { Then } a b^{5}+b c^{5}+c a^{5} \\
=-\frac{1}{a^{4} c^{5}}-\frac{c^{4}}{a}+c a^{5}=\frac{a^{9} c^{6}-1-a^{3} c^{9}}{a^{4} c^{5}} \\
=\frac{\left(a c^{3}+1\right)^{3}-1-a^{3} c^{9}}{a^{4} c^{5}}=\frac{3\left(a^{2} c^{6}+a c^{3}\right)}{a^{4} c^{5}} \\
=\frac{3\left(a c^{3}+1\right)}{a^{3} c^{2}}=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1. Then
$\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=$ $\qquad$ [2]
(2012, Zhejiang Province High School Mathematics Competition)
|
When the common ratio $q=1$,
$$
\begin{array}{l}
a_{n}=a_{1}, \\
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011 .
\end{array}
$$
When the common ratio $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} \sum_{i=1}^{2011}\left(\frac{1}{\lg a_{i}}-\frac{1}{\lg a_{i+1}}\right) \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q}\left(\frac{1}{\lg a_{1}}-\frac{1}{\lg a_{2012}}\right) \\
=2011 .
\end{array}
$$
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfy
$$
\frac{a_{j}}{a_{i}}>\frac{2}{3}(1 \leqslant i \leqslant j \leqslant 10) \text {. }
$$
Then the minimum possible value of $a_{10}$ is $\qquad$ .
|
2. 92.
From $a_{1} \geqslant 1, a_{2}>\frac{3}{2} a_{1} \geqslant \frac{3}{2}\left(a_{2} \in \mathbf{N}_{+}\right)$, we get $a_{2} \geqslant 2$.
Similarly, $a_{3}>\frac{3}{2} a_{2} \geqslant 3, a_{3} \geqslant 4$;
$$
\begin{array}{l}
a_{4}>\frac{3}{2} a_{3} \geqslant 6, a_{4} \geqslant 7 ; \\
a_{5}>\frac{3}{2} a_{4} \geqslant \frac{21}{2}, a_{5} \geqslant 11 ;
\end{array}
$$
$$
\begin{array}{l}
a_{6}>\frac{3}{2} a_{5} \geqslant \frac{33}{2}, a_{6} \geqslant 17 ; \\
a_{7}>\frac{3}{2} a_{6} \geqslant \frac{51}{2}, a_{7} \geqslant 26 ; \\
a_{8}>\frac{3}{2} a_{7} \geqslant 39, a_{8} \geqslant 40 ; \\
a_{9}>\frac{3}{2} a_{8} \geqslant 60, a_{9} \geqslant 61 ; \\
a_{10}>\frac{3}{2} a_{9} \geqslant \frac{183}{2}, a_{10} \geqslant 92 .
\end{array}
$$
Therefore, the smallest possible value of $a_{10}$ is 92.
|
92
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$. Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$
|
3. 11 .
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then the given equations can be written as
$$
\begin{array}{l}
x+y+z=\frac{17}{6}, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}, \\
\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5} . \\
\text { From (1) } \div \text { (3) we get } \\
x y z=-\frac{5}{6} .
\end{array}
$$
From (1) $\div$ (3) we get
From (2) $\times$ (4) we get $x y+y z+z x=\frac{2}{3}$.
Thus $\tan (\alpha+\beta+\gamma)=\frac{x+y+z-x y z}{1-(x y+y z+z x)}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$
If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$
|
8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l}
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \frac{1}{3}\left(a_{2}-a_{1}\right) \\
= \frac{1 \times 2}{n(n-1)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
\Rightarrow a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=1+2\left(1-\frac{1}{n}\right)=3-\frac{2}{n} .
\end{array}
\end{array}
$$
Thus, $a_{m}>2+\frac{2011}{2012}$
$$
\begin{array}{l}
\Leftrightarrow 3-\frac{2}{m}>3-\frac{1}{2012} \\
\Leftrightarrow m>4024 .
\end{array}
$$
In summary, the smallest value of $m$ that satisfies the condition is 4025.
|
4025
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) Given an integer $n(n \geqslant 3)$, let $f(n)$ be the minimum number of elements in a subset $A$ of the set $\left\{1,2, \cdots, 2^{n}-1\right\}$ that satisfies the following two conditions:
(i) $1 \in A, 2^{n}-1 \in A$;
(ii) Each element in subset $A$ (except 1) is the sum of two (possibly the same) elements in $A$.
(1) Find the value of $f(3)$;
(2) Prove: $f(100) \leqslant 108$.
|
12. (1) Let set $A \subseteq\left\{1,2, \cdots, 2^{3}-1\right\}$, and $A$ satisfies (i) and (ii). Then $1 \in A, 7 \in A$.
Since $\{1, m, 7\}(m=2,3, \cdots, 6)$ does not satisfy (ii), hence $|A|>3$.
Also, $\{1,2,3,7\},\{1,2,4,7\},\{1,2,5,7\}$,
$\{1,2,6,7\},\{1,3,4,7\},\{1,3,5,7\}$,
$\{1,3,6,7\},\{1,4,5,7\},\{1,4,6,7\}$,
$\{1,5,6,7\}$
do not satisfy (ii), hence $|A|>4$.
However, the set $\{1,2,4,6,7\}$ satisfies (i) and (ii), so $f(3)=5$.
(2) First, prove:
$$
f(n+1) \leqslant f(n)+2(n=3,4, \cdots) \text {. }
$$
In fact, if $A \subseteq\left\{1,2, \cdots, 2^{n}-1\right\}$ satisfies (i) and (ii), and the number of elements in set $A$ is $f(n)$. Let
$$
B=A \cup\left\{2^{n+1}-2,2^{n+1}-1\right\} \text {. }
$$
Since $2^{n+1}-2>2^{n}-1$, we have
$|B|=f(n)+2$.
Also, $2^{n+1}-2=2\left(2^{n}-1\right)$,
$$
2^{n+1}-1=1+\left(2^{n+1}-2\right) \text {, }
$$
then $B \subseteq\left\{1,2, \cdots, 2^{n+1}-1\right\}$, and set $B$ satisfies
(i) and (ii).
Thus, $f(n+1) \leqslant|B|=f(n)+2$.
Next, prove:
$$
f(2 n) \leqslant f(n)+n+1(n=3,4, \cdots) \text {. }
$$
In fact, let $A \subseteq\left\{1,2, \cdots, 2^{n}-1\right\}$ satisfy
(i) and (ii), and the number of elements in set $A$ is $f(n)$. Let
$$
\begin{aligned}
B= & A \cup\left\{2\left(2^{n}-1\right), 2^{2}\left(2^{n}-1\right), \cdots,\right. \\
& \left.2^{n}\left(2^{n}-1\right), 2^{2 n}-1\right\} .
\end{aligned}
$$
Since $2\left(2^{n}-1\right)<2^{2}\left(2^{n}-1\right)<\cdots$
$$
<2^{n}\left(2^{n}-1\right)<2^{2 n}-1 \text {, }
$$
then $B \subseteq\left\{1,2, \cdots, 2^{2 n}-1\right\}$, and
$$
\begin{array}{l}
|B|=f(n)+n+1 . \\
\text { and } 2^{k+1}\left(2^{n}-1\right) \\
=2^{k}\left(2^{n}-1\right)+2^{k}\left(2^{n}-1\right)(k=0,1, \cdots, n-1), \\
2^{2 n}-1=2^{n}\left(2^{n}-1\right)+\left(2^{n}-1\right),
\end{array}
$$
then $B$ satisfies (i) and (ii).
Thus, $f(2 n) \leqslant|B|=f(n)+n+1$.
From equations (1) and (2), we get
$$
f(2 n+1) \leqslant f(n)+n+3 \text {. }
$$
Repeatedly using equations (2) and (3), we get
$$
\begin{array}{l}
f(100) \leqslant f(50)+51 \\
\leqslant f(25)+26+51 \\
\leqslant f(12)+15+77 \\
\leqslant f(6)+7+92 \\
\leqslant f(3)+4+99=108 .
\end{array}
$$
(Xiong Hua, Gu Hongda, Liu Hongkuang, Li Dayuan, Ye Shengyang)
|
108
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, c$ satisfy
$$
a+b+c=a^{2}+b^{2}+c^{2} \text {. }
$$
Then the maximum value of $a+b+c$ is $\qquad$
|
4.3.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
3(a+b+c) \\
=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
$$
\begin{array}{l}
\geqslant(a+b+c)^{2} \\
\Rightarrow a+b+c \leqslant 3 .
\end{array}
$$
Equality holds if and only if \(a=b=c=1\).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Arrange the positive integers whose sum of digits is 5 in ascending order to form a sequence. Then 2012 is the $\qquad$th term of this sequence.
|
7.38.
To represent 5 as the sum of no more than four positive integers, there are six methods, that is
$$
\begin{array}{l}
5=1+4=2+3=1+1+3 \\
=1+2+2=1+1+1+2 .
\end{array}
$$
When filling them into a $1 \times 4$ grid, positions that are not filled are supplemented with 0. Then $\{5\}$ has 3 ways of filling; $\{1,4\}$ has 9 ways of filling; $\{2,3\}$ has 7 ways of filling; $\{1,1,3\}$ has 9 ways of filling; $\{1,2,2\}$ has 7 ways of filling; $\{1,1,1,2\}$ has 3 ways of filling.
There are a total of 38 ways of filling, with 2012 being the largest.
Therefore, 2012 is the 38th term.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$
|
8.5476.
For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} \pm 40 a+4 .
\end{array}
$$
If following the above rules, it can only be the case of
$$
3456,4536,5436,5364,3564 \text {. }
$$
But none of the above are square numbers.
The next four consecutive digits are $4,5,6,7$.
If arranged into a square number, the last digit can only be 4 or 6, from smallest to largest:
$$
4756,5476,5764,7564 .
$$
After calculation, only $5476=74^{2}$ is a square number.
|
5476
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$. Then $a^{b}=$ $\qquad$ .
|
- 1. 16.
From $\log _{2} a=\frac{a+1}{b} \Rightarrow a^{b}=2^{a+1}$.
Substituting into $4^{a}-3 a^{b}=16$, we get
$2^{2 a}-6 \times 2^{a}-16=0$.
Solving, we get $2^{a}=8$ or -2 (discard).
Therefore, $a^{b}=2^{a+1}=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the base edge length of a regular tetrahedron is 6, and the side edge is 4. Then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$
|
3. 4 .
From the problem, we know that the distance from point $A$ to the base $B C D$ is 2, and the radius of its circumscribed sphere is $R$. Then
$$
R^{2}-12=(R-2)^{2} \Rightarrow R=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $S_{4} \leqslant 4, S_{5} \geqslant 15$, then the minimum value of $a_{4}$ is $\qquad$ .
|
4. 7 .
Let the common difference be $d$. From the given conditions, we have
$$
\begin{array}{l}
2 a_{4} \leqslant 2+3 d, d \leqslant a_{4}-4 \\
\Rightarrow 2 a_{4} \leqslant 2+3 d \leqslant 2+3\left(a_{4}-3\right) \\
\Rightarrow a_{4} \geqslant 7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
the value.
|
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos \frac{B}{2} \cdot \cos \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2}\left(\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)^{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
=\left(\frac{a+b+c}{2}\right)^{2}=64 .
\end{array}
$$
Solution 2: As shown in Figure 2, extend line segment $BC$ to points $E$ and $F$ such that $CE = AC = b$ and $BF = AB = c$. Connect $AE$ and $AF$. Let the midpoints of $AE$ and $AF$ be $M$ and $N$, respectively. Connect $BN$, $CM$, and $NM$.
$$
\begin{array}{l}
\text { Then } AM = b \cos \frac{C}{2}, AN = c \cos \frac{B}{2}, \\
\angle MAN = \angle A + \frac{1}{2}(\angle B + \angle C) = \frac{\pi}{2} + \frac{\angle A}{2}. \\
\text { Therefore, } b^{2} \cos ^{2} \frac{C}{2} + c^{2} \cos ^{2} \frac{B}{2} + 2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= AM^{2} + AN^{2} - 2 AM \cdot AN \cos \angle MAN \\
= MN^{2} = \left(\frac{a+b+c}{2}\right)^{2} = 64 .
\end{array}
$$
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, it is known that $BE \perp AC$ at point $E$, $CD \perp AB$ at point $D$, $BC=25$, $CE=7$, $BD=15$. If $BE$ and $CD$ intersect at point $H$, connect $DE$, and construct a circle with $DE$ as the diameter, which intersects $AC$ at another point $F$. Find the length of $AF$.
(2012, Joint Autonomous Admissions Examination of Tsinghua University and Other Universities)
|
Solve for DF. From the given conditions, we have
$$
\begin{array}{l}
\cos B=\frac{3}{5}, \cos C=\frac{7}{25} \\
\Rightarrow \sin B=\frac{4}{5}, \sin C=\frac{24}{25} \\
\Rightarrow \sin A=\sin B \cdot \cos C+\sin C \cdot \cos B=\frac{4}{5}=\sin B \\
\Rightarrow \angle A=\angle B \Rightarrow AC=BC=25 .
\end{array}
$$
From $CD \perp AB, DF \perp AC$, we know that $D$ and $F$ are the midpoints of $AB$ and $AE$, respectively.
Also, $AE=AC-CE=18$, so $AF=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the left and right foci respectively, and $O$ be the center. Then
$\left|P F_{1}\right|\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ .
|
16. 25.
According to the definition of an ellipse and the cosine rule, the solution can be found.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let two ellipses be
$$
\frac{x^{2}}{t^{2}+2 t-2}+\frac{y^{2}}{t^{2}+t+2}=1
$$
and $\frac{x^{2}}{2 t^{2}-3 t-5}+\frac{y^{2}}{t^{2}+t-7}=1$
have common foci. Then $t=$ $\qquad$ .
|
5.3.
Given that the two ellipses have a common focus, we have
$$
\begin{array}{l}
t^{2}+2 t-2-\left(t^{2}+t+2\right) \\
=2 t^{2}-3 t-5-\left(t^{2}+t-7\right) \\
\Rightarrow t=3 \text { or } 2 \text { (rejected). }
\end{array}
$$
Therefore, $t=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 (An Ancient Chinese Mathematical Problem) Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people.
|
Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people.
From the equality of the total number of soldiers, we get the equation
$$
1001 x + 1 = 1002 y + 4 \text{. }
$$
Rearranging gives
$$
1001(x - y) - 3 = y \text{. }
$$
Since $x$ and $y$ are both positive integers, the smallest value for $x - y$ is 1, at which point, $y = 998$.
Thus, the minimum number of soldiers in this troop count is
$$
998 \times 1002 + 4 = 1000000 \text{ (people). }
$$
|
1000000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$
|
5. 0 .
Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}}$.
Perform a summation index transformation $l=2 n-k-1$, then
$$
\begin{array}{l}
a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \\
=\sum_{l=0}^{2 n-1}(-1)^{\prime} \frac{l+1}{\mathrm{C}_{2 n}^{l}}=-a_{n} . \\
\text { Hence } a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let $a, b, c > 0$, and $a+b+c=3$. Find the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$.
|
Let's assume $a \leqslant b \leqslant c$ or $c \leqslant b \leqslant a$.
Then $c(b-a)(b-c) \leqslant 0$
$$
\begin{aligned}
\Rightarrow & b^{2} c+c^{2} a \leqslant a b c+c^{2} b \\
\Rightarrow & a^{2} b+b^{2} c+c^{2} a+a b c \\
& \leqslant a^{2} b+c^{2} b+2 a b c \\
& =b(a+c)^{2}=b(3-b)^{2} \\
& \leqslant \frac{1}{2}\left(\frac{2 b+3-b+3-b}{3}\right)^{3}=4 .
\end{aligned}
$$
Equality holds if and only if $a=b=c=1$ or $\{a, b, c\}=\{0, 1, 2\}$.
Thus, the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1.
|
If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.
If there exists a point (let's assume it is $v$) with a degree less than or equal to 3, then at least four points not connected to $v$ must be pairwise connected, which is clearly impossible.
Therefore, the degree of any point is at least 4.
In the point set $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$, take a point (let's assume it is $v$) at the vertex of the convex hull $T$. Point $v$ is connected to $v_{1}, v_{2}, v_{3}, v_{4}$ (where $v_{1}, v_{2}, v_{3}, v_{4}$ are arranged in a counterclockwise direction). Then $\left|v_{1} v_{4}\right| \neq 1$ (otherwise, $v_{2}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction). Therefore, $v_{2}, v_{3}$ must each be connected to at least one of $v_{1}, v_{4}$. Thus, $v_{2}, v_{3}$ are each connected to exactly one of $v_{1}, v_{4}$ (otherwise, assume $v_{2}$ is connected to both $v_{1}, v_{4}$, then $v_{1}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction).
Moreover, $v_{2}, v_{3}$ cannot both be connected to the same point among $v_{1}, v_{4}$. Therefore, among the points $v_{1}, v_{2}, v_{3}, v_{4}$, only $\left|v_{1} v_{2}\right|=\left|v_{3} v_{4}\right|=1$ or $\left|v_{1} v_{3}\right|=\left|v_{2} v_{4}\right|=1$ (with $\left.\left|v_{2} v_{3}\right| \neq 1\right)$ can occur.
Thus, each of $v_{1}, v_{2}, v_{3}, v_{4}$ must be connected to at least two of $v_{5}, v_{6}, v_{7}$. Therefore, the points $v_{5}, v_{6}, v_{7}$ must draw at least eight edges to $v_{1}, v_{2}, v_{3}, v_{4}$. By the pigeonhole principle, at least one point (let's assume it is $v_{5}$) must draw at least three edges to $v_{1}, v_{2}, v_{3}, v_{4}$, meaning there exist three points among $v_{1}, v_{2}, v_{3}, v_{4}$ that are all 1 unit away from both $v$ and $v_{5}$, which is impossible.
Therefore, $n \leqslant 7$.
Take two rhombuses $A B C D$ and $A B_{1} C_{1} D_{1}$ with side lengths of 1, such that
$$
\angle B A D=\angle B_{1} A D_{1}=60^{\circ}, C C_{1}=1 .
$$
It is easy to prove that among $A, B, C, D, B_{1}, C_{1}, D_{1}$, any three points have at least two points whose distance is 1.
Therefore, the maximum value of $n$ is 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $x, y$ are positive real numbers, $n \in \mathrm{N}$, and $n \geqslant 2$. Prove:
$$
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \geqslant 4 \text {. }
$$
|
$$
\begin{array}{l}
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \\
\geqslant 2 \sqrt{\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}} \cdot \sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}}} \\
=2 \sqrt[2 n]{\frac{\left(2^{n}-1\right)\left(x^{2}+y^{2}\right)+\left(2^{2 n}-2^{n+1}+2\right) x y}{x y}} \\
\geqslant 2 \sqrt[2 n]{\frac{2\left(2^{n}-1\right) x y+\left(2^{2 n}-2^{n+1}+2\right) x y}{x y}} \\
=4 .
\end{array}
$$
The equality holds if and only if \( x = y \).
|
4
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Mother's Day is coming, and Xiao Hong, Xiao Li, and Xiao Meng went to the flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the three types of flowers. Then she should pay $\qquad$ yuan. $[1]$
|
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $\sqrt{z}$ yuan, respectively. Then,
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16 .
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text{. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text{. }
$$
From equations (2) and (3), we get
$$
x+y+z=10 \text{, }
$$
which means $2(x+y+z)=20$.
Therefore, Xiaoying should pay 20 yuan.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$
and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$
|
Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$
Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \angle C B D=\angle C A D=\angle A D B, \\
A B=C D .
\end{array}
$$
Take the midpoints $E$ and $F$ of $C D$ and $A B$ respectively, and connect $A E$, $B E$, and $E F$.
Then $A E \perp C D, B E \perp C D$.
Thus, $C D \perp$ plane $A B E$.
Therefore, plane $A B E$ is the perpendicular plane of $C D$.
It is calculated that $A E=B E=9 \sqrt{2}, E F=12$. $=\frac{1}{3} \times \frac{1}{2} \times 6 \sqrt{2} \times 12 \times 6 \sqrt{2}=144$.
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let the functions $f(x)=\ln x, g(x)=\frac{1}{2} x^{2}$. If $x_{1}>x_{2}>0$, for what value of $m(m \in \mathbf{Z}, m \leqslant 1)$ is it always true that
$$
m\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)>x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)
$$
holds.
|
Introduce an auxiliary function
$$
t(x)=m g(x)-x f(x)=\frac{m}{2} x^{2}-x \ln x \quad (x>0) \text {. }
$$
By the problem, $x_{1}>x_{2}>0$.
Therefore, if the original inequality always holds for $x>0$, i.e., the function $t(x)$ is monotonically increasing, then
$$
t^{\prime}(x)=m x-\ln x-1 \geqslant 0
$$
always holds.
Thus, $m \geqslant \frac{\ln x+1}{x}$ always holds, which means
$$
m \geqslant\left(\frac{\ln x+1}{x}\right)_{\max } \text {. }
$$
Introduce another auxiliary function
$$
h(x)=\frac{\ln x+1}{x} \quad (x>0) \text {. }
$$
By $h^{\prime}(x)=\frac{-\ln x}{x^{2}}$, we know that the function $h(x)$ is monotonically increasing on $(0,1]$ and monotonically decreasing on $(1,+\infty)$, i.e.,
$$
h(x)_{\max }=h(1)=1 \quad (m \geqslant 1) .
$$
Since $m \in \mathbf{Z}, m \leqslant 1$, hence $m=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had at least $\qquad$ pieces of candy.
|
Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$
Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$
which simplifies to $x=\frac{160 m+15}{m+1}=160-\frac{145}{m+1}$.
Since $x$ and $m$ are both positive integers, we have:
$$
\begin{array}{l}
160-\frac{29 \times 5}{m+1} \\
\geqslant 160-29=131(m>1) .
\end{array}
$$
Therefore, the first box originally contained at least 131 candies.
|
131
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$
the solution is $1 \leqslant x \leqslant 3$. Then $a=$
|
$=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$.
|
-12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x$ and $y$ be two distinct non-negative integers, and satisfy $x y + 2x + y = 13$. Then the minimum value of $x + y$ is $\qquad$
|
3. 5 .
From the problem, we know that $(x+1)(y+2)=15$.
$$
\begin{array}{l}
\text { Then }(x+1, y+2) \\
=(15,1),(5,3),(3,5),(1,15) \\
\Rightarrow(x, y)=(14,-1),(4,1),(2,3),(0,13) .
\end{array}
$$
Therefore, the minimum value of $x+y$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given three points $A, B, C$ in a plane satisfying
$$
|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=5,|\overrightarrow{C A}|=6 \text {. }
$$
Then the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is ( ).
|
5. C.
From the cosine theorem, we know
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-\overrightarrow{B A} \cdot \overrightarrow{B C} \\
=-\frac{|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}-|\overrightarrow{A C}|^{2}}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\overrightarrow{B C} \cdot \overrightarrow{C A}=-\frac{|\overrightarrow{C A}|^{2}+|\overrightarrow{C B}|^{2}-|\overrightarrow{A B}|^{2}}{2}, \\
\overrightarrow{C A} \cdot \overrightarrow{A B}=-\frac{|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}-|\overrightarrow{B C}|^{2}}{2} .
\end{array}
$$
Therefore, $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$
$$
=-\frac{|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+|\overrightarrow{B C}|^{2}}{2}=-35 \text {. }
$$
|
-35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$
Then $1992 a c-1811 b c-181 a b=$
|
12. 0 .
Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$
Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac{a-c}{d}+\frac{c-b}{d}+\frac{b-a}{d}=0 .
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$
|
13. 1 .
Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$
we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$
Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}\right| \leqslant 1,\left|\sin \frac{a-b}{2}\right|<\frac{a-b}{2} \text {. }
$$
Thus $|a-b|<2\left|\frac{a-b}{2}\right|=|a-b|$, which is a contradiction.
Since the graphs of $y=x$ and $y=\cos (x+c)$ must intersect, we know that $a=b$ holds.
Therefore, $\log _{a} b=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Every day at 5 PM when school is over, Xiao Ming's father always drives from home to pick him up on time and take him back. One day, the school dismissed an hour early, and Xiao Ming walked home by himself. On the way, he met his father who was coming to pick him up, and as a result, they arrived home 20 minutes earlier than usual. Then Xiao Ming walked for $\qquad$ minutes before meeting his father.
|
二、1.50 minutes.
As shown in Figure 6, Xiao Ming
starts walking home from point $A$
and meets the car coming to pick him up
at point $C$. As a result, the car returns from $C$ to $B$ 20 minutes earlier than usual. This indicates that the car takes 20 minutes to travel from $C$ to $A$ and back to $C$.
Therefore, the car takes 10 minutes to travel from $C$ to $A$.
Thus, the car meets Xiao Ming at $4:50$, meaning Xiao Ming has been walking for 50 minutes when he meets his father.
|
50
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$
|
2. 45 .
From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$
When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\
\Rightarrow x+y+z+w \geqslant 20 .
\end{array}
$$
When $x=20, y=z=w=0$, the equality holds. Therefore, $a=25, b=20$.
Thus, $a+b=45$.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$
has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$
has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$
has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$
|
3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$
Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) In a dormitory of a school, there are several students. On New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator, who in turn gives a card back to each student. In this way, a total of 51 greeting cards were used. Question: How many students live in the dormitory?
|
Let there be $x$ students and $y$ administrators living in the dormitory $\left(x, y \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
x(x-1)+x y+y=51 \\
\Rightarrow y=\frac{51+x-x^{2}}{x+1}=\frac{49}{x+1}-x+2 .
\end{array}
$$
Since $x, y$ are positive integers, we know that $(x, y)=(6,3)$.
Therefore, there are 6 students living in the dormitory.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, it is known that $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=$ $\qquad$
|
3. 0 .
It is known that, $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O H}$.
Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{M N}$
$$
\begin{aligned}
= & \overrightarrow{O A} \cdot(\overrightarrow{M F}+\overrightarrow{F E}+\overrightarrow{E N})+ \\
& \overrightarrow{O B} \cdot(\overrightarrow{M F}+\overrightarrow{F N})+\overrightarrow{O C} \cdot(\overrightarrow{M E}+\overrightarrow{E N}) \\
= & (\overrightarrow{O A}+\overrightarrow{O B}) \cdot \overrightarrow{M F}+(\overrightarrow{O A}+\overrightarrow{O C}) \cdot \overrightarrow{E N} .
\end{aligned}
$$
Since $O$ is the circumcenter of $\triangle A B C$, $\overrightarrow{O A}+\overrightarrow{O B}$ and $\overrightarrow{O A}+\overrightarrow{O C}$ pass through the midpoints of segments $A B$ and $A C$, respectively.
Therefore, the value of equation (1) is 0.
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$
|
$\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{array}$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of five-digit numbers that remain the same after a $180^{\circ}$ rotation is.
|
2. 60.
Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. Such five-digit numbers include: 18 without $(6,9)$; exactly one pair of $(6,9)$ has 30; two pairs of $(6,9)$ have 12. Therefore, the total number of such five-digit numbers is 60.
|
60
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, in
rhombus $A B C D$, it is
known that $\angle A B C=60^{\circ}$, line
$E F$ passes through point $D$, and
intersects the extensions of
$B A$ and $B C$ at points $E$ and $F$, respectively. $M$
is the intersection of $C E$ and $A F$. If $C M=4, E M=5$, then $C A=$ $\qquad$
|
3. 6 .
It is easy to prove $\triangle E A D \backsim \triangle D C F$.
$$
\begin{array}{l}
\text { Therefore, } \frac{E A}{A D}=\frac{D C}{C F} \Rightarrow \frac{E A}{A C}=\frac{A C}{C F} \\
\Rightarrow \triangle E A C \backsim \triangle A C F \\
\Rightarrow \angle A E C=\angle C A F \\
\Rightarrow C A^{2}=C E \cdot C M=9 \times 4 \\
\Rightarrow C A=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers.
|
Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Rightarrow a_{9}=a_{8}+d_{8} \geqslant 18 \\
\Rightarrow d_{9} \geqslant \frac{18^{2}}{31-18}>24 \Rightarrow d_{9} \geqslant 25 \\
\Rightarrow a_{10}=a_{9}+d_{9} \geqslant 43>31 .
\end{array}
$$
Therefore, $n-1 \leqslant 9 \Rightarrow n \leqslant 10$.
At the same time, the ten natural numbers $1,2,3,4,5,6,8, 11,18, 43$ meet the conditions of the problem.
Thus, the maximum value of $n$ is 10.
|
10
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the sum of the first $n(n>1)$ terms of an arithmetic sequence is 2013, the common difference is 2, and the first term is an integer. Then the sum of all possible values of $n$ is . $\qquad$
|
3. 2975 .
Let the first term of the sequence be $a_{1}$, and the common difference $d=2$. Then
$$
\begin{array}{l}
S_{n}=n a_{1}+n(n-1) \\
=n\left(a_{1}+n-1\right)=2013 .
\end{array}
$$
Also, $2013=3 \times 11 \times 61$, and $n$ is a divisor of 2013, so the sum of all possible values of $n$ is
$$
(1+3) \times(1+11) \times(1+61)-1=2975 .
$$
|
2975
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a, b, x \in \mathbf{N}_{+}$, and $a \leqslant b$. $A$ is the solution set of the inequality
$$
\lg b - \lg a < \lg x < \lg b + \lg a
$$
It is known that $|A|=50$. When $ab$ takes its maximum possible value,
$$
\sqrt{a+b}=
$$
|
4. 6 .
It is easy to know, $\frac{b}{a}<x<a b, a \neq 1$.
Therefore, $a \geqslant 2$,
$$
\begin{array}{l}
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 \\
\Rightarrow a b \leqslant 68 .
\end{array}
$$
Upon inspection, when and only when $a=2, b=34$, the equality holds.
At this time, $\sqrt{a+b}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function defined on the domain $R$
$$
f(x)=|\lg | x-2||-1 \text{. }
$$
If $b<0$, then the equation concerning $x$
$$
f^{2}(x)+b f(x)=0
$$
has $\qquad$ distinct real roots.
|
5. 8 .
From the problem, we know that the graph of $y=\lg x$ is first symmetrical about the $y$-axis, then the part below the $x$-axis is flipped up, followed by a rightward shift of 2 units and a downward shift of 1 unit to form the graph of $f(x)$. The zeros of $g(x)=f^{2}(x)+b f(x)$ are the roots of the equation. It is easy to see that $g(x)$ has 8 zeros.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$
where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$.
|
10. Let $b_{n}=a_{n+1}-a_{n}$.
From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $n \geqslant 3$, and
$$
\begin{array}{l}
\left\{\begin{array}{l}
a_{n} \leqslant a_{n+1}, \\
a_{n} \leqslant a_{n-1},
\end{array}\right. \\
\left\{\begin{array}{l}
b_{n}=\frac{(n-1)(n-40)}{2}+1 \geqslant 0, \\
b_{n-1}=\frac{(n-2)(n-41)}{2}+1 \leqslant 0 .
\end{array}\right.
\end{array}
$$
Therefore, when $n=40$, the value of $a_{n}$ is the smallest.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.
Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}-1$.
Hence $51117^{2 y-1} \Rightarrow 7317^{2 y-2}$, which is a contradiction.
Therefore, $z \geqslant 3 \times 7 \times 8+1=169$.
Let $z=169$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=169$.
Solving this, we get $(x, y)=(1,2)$.
Thus, the smallest non-negative value of $z$ is 169.
(2) $x=2 x_{1}$ is an even number.
Then $z \equiv(-1)^{x}-1 \equiv 0(\bmod 3)$.
Thus, $z \equiv 57(\bmod 3 \times 7 \times 8)$.
Let $z=57$, i.e., $512^{2 x_{1}}-7^{2 y-1}=57$.
Let $512^{x_{1}}=s, 7^{y-1}=t$. Then $s^{2}-7 t^{2}=57$.
The fundamental solution of the equation $s^{2}-7 t^{2}=1$ is $8+3 \sqrt{7}$.
Let $s_{1}+t_{1} \sqrt{7}$ be the fundamental solution of $s^{2}-7 t^{2}=57$. Then all solutions of the equation are
$$
S_{n+1}+t_{n+1} \sqrt{7}=\left(s_{n}+t_{n} \sqrt{7}\right)(8+3 \sqrt{7}),
$$
i.e.,
$$
s_{n+1}=8 s_{n}+21 t_{n}, t_{n+1}=3 s_{n}+8 t_{n}.
$$
Solving the first equation for $t_{n}=\frac{s_{n+1}-8 s_{n}}{21}$ and substituting into the second equation, we get
$$
\begin{array}{l}
s_{n+2}=16 s_{n+1}-s_{n} \\
\Rightarrow s_{n+2}=-s_{n}(\bmod 16).
\end{array}
$$
The fundamental solution $s_{1}+t_{1} \sqrt{7}$ should satisfy
$$
\begin{array}{l}
0 \leqslant s_{1} \leqslant \sqrt{\frac{1}{2}(8+1) \times 57}, \\
0 \leqslant t_{1} \leqslant \frac{3 \sqrt{27}}{\sqrt{2(8+1)}},
\end{array}
$$
i.e., $0 \leqslant s_{1} \leqslant 16,0 \leqslant t_{1} \leqslant 5$.
Trying each one, we find only
$$
\left(s_{1}, t_{1}\right)=(8,1),(13,4).
$$
When $\left(s_{1}, t_{1}\right)=(8,1)$, by conclusion (1) we get
$$
s_{2}=85 \equiv 5(\bmod 16);
$$
When $\left(s_{1}, t_{1}\right)=(13,4)$, by conclusion (1) we get $s_{2}=188 \equiv 12(\bmod 16)$.
Thus, $s_{n} \neq 0(\bmod 16)$, which is a contradiction.
Therefore, $512^{2 x_{1}}-7^{2 y-1}=57$ has no positive integer solutions.
Thus, $512^{2 x_{1}}-7^{2 y-1} \geqslant 168+57$.
In summary, the minimum value of $\sqrt{512^{x}-7^{2 y-1}}$ is 13.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $z$ be a complex number with modulus 2. Then the sum of the maximum and minimum values of $\left|z-\frac{1}{z}\right|$ is $\qquad$ [2]
|
Given $|z|=2$, we know
$$
\begin{array}{l}
|z+1|^{2}=(z+1)(\bar{z}+1) \\
=z \bar{z}+z+\bar{z}+1=5+2 \operatorname{Re} z, \\
|z-1|^{2}=(z-1)(\bar{z}-1) \\
=z \bar{z}-z-\bar{z}+1=5-2 \operatorname{Re} z .
\end{array}
$$
Therefore, $\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|$
$$
=\frac{|z+1||z-1|}{|z|}=\frac{\sqrt{25-4 \operatorname{Re} z}}{2} \text {. }
$$
Its maximum value is $\frac{5}{2}$, and the minimum value is $\frac{3}{2}$.
Thus, the required result is 4.
1.3 Roots of Unity in Complex Numbers
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given
$$
A=\left\{z \mid z^{18}=1\right\} \text { and } B=\left\{\omega \mid \omega^{48}=1\right\}
$$
are sets of complex roots of unity,
$$
C=\{z w \mid z \in A, w \in B\}
$$
is also a set of complex roots of unity. How many distinct elements are there in the set $C$? ${ }^{[3]}$
|
Notice that, $z=\cos \frac{2 k \pi}{18}+\mathrm{i} \sin \frac{2 k \pi}{18}(k \in \mathbf{Z})$ (18 distinct elements),
$\omega=\cos \frac{2 t \pi}{48}+\mathrm{i} \sin \frac{2 t \pi}{48}(t \in \mathbf{Z})$ (48 distinct elements),
$$
\begin{array}{l}
z \omega=\cos \frac{2 \pi(8 k+3 t)}{144}+\mathrm{i} \sin \frac{2 \pi(8 k+3 t)}{144} . \\
\text { Let } P=\{m \mid m=8 k+3 t, t \in \mathbf{Z}\} .
\end{array}
$$
By Bézout's theorem, $P=\mathbf{Z}$.
Therefore, the set $C$ contains 144 distinct elements.
|
144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$
|
Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mathrm{i}\right]\right| \\
=\left|n^{2}+17 \mathrm{i}\right|=\sqrt{n^{4}+17^{2}}
\end{array}
$$
From the given condition, we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}\left(m \in \mathbf{N}_{+}, m=S_{n}\right) \\
\Rightarrow\left(m-n^{2}\right)\left(m+n^{2}\right)=289 \\
\Rightarrow m-n^{2}=1, m+n^{2}=289 \\
\Rightarrow n=12 .
\end{array}
$$
2.2 Application in Trigonometric Problems
The connection between complex numbers and trigonometric functions mainly relies on the exponential form (or trigonometric form) of complex numbers. By using the exponential form and operations (or trigonometric operations) of complex numbers, we can achieve the purpose of trigonometric evaluation and proof.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In rectangle $A B C D$, it is known that $A B=5, B C=9$, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=C G=3, B F=D H=4, P$ is a point inside the rectangle. If the area of quadrilateral $A E P H$ is 15, then the area of quadrilateral $P F C G$ is $\qquad$
|
4. 11.
As shown in Figure 3, let the distances from $P$ to $AB$ and $AD$ be $a$ and $b$, respectively. Then the distances from $P$ to $BC$ and $CD$ are $5-b$ and $9-a$, respectively. Given that $S_{\text {quadrilateral } AEPH}=\frac{1}{2}(3a+5b)=15$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral PFCG }}=\frac{1}{2}[5(5-b)+3(9-a)] \\
=\frac{1}{2}[52-(3a+5b)]=11 .
\end{array}
$$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5.4.
If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$
Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (corresponding to $n^{4}-3 n^{2}+9=7$ or 13, both of which are prime numbers).
If $n<0$, then $n=-1$ or -2.
In summary, there are 4 integers $n$ that satisfy the condition.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$
|
7. 100 .
Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text { gives } \\
6 n-60=0 \Rightarrow n=10 .
\end{array}
$$
Therefore, the area of the trapezoid is
$$
\frac{1}{2}(9+11) \times 10=100 \text {. }
$$
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer not exceeding the real number $x$).
|
8. 2025078.
Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$
From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$
Therefore, $\left[S_{2 n}\right]=2 n$.
Also, $S_{2 n-1}=S_{2 n}-(4 n-1)=4 n^{2}-3 n+1$, then $(2 n-1)^{2}<S_{2 n-1}<4 n^{2}$.
Therefore, $\left[S_{2 n-1}\right]=2 n-1$.
$$
\begin{array}{l}
\text { Hence }\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right] \\
=1+2+\cdots+2012=2025078 .
\end{array}
$$
|
2025078
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 A scientist stored the design blueprint of his time machine in a computer, setting the password to open the file as a permutation of $\{1,2, \cdots, 64\}$. He also designed a program that, when eight positive integers between $1 \sim 64$ are input each time, the computer will indicate the order (from left to right) of these eight numbers in the password. Please design an operation scheme such that the password can be determined with at most 45 inputs. [6]
|
Let the password be denoted as $a_{1} a_{2} \cdots a_{n^{2}} (n=8)$.
First, write the numbers $1, 2, \cdots, n^{2}$ arbitrarily into an $n \times n$ grid (one number per cell). After the first $n$ operations, input each row of numbers once, and rearrange the numbers in each row from left to right according to the computer's prompt.
Color the numbers in the 1st and $n$-th columns red and yellow, respectively, and color the numbers in the other columns blue.
In the $(n+1)$-th and $(n+2)$-th operations, input the numbers in the 1st and $n$-th columns, respectively, to determine the order of the red and yellow numbers in the password, and identify $a_{1}$ and $a_{n^{2}}$. Remove these two numbers from the 1st and $n$-th columns, and replace them with adjacent blue numbers.
In the $(n+3)$-th operation, input the $\frac{n}{2}-1$ leftmost (rightmost) red (yellow) numbers in the password along with the blue numbers in the 1st and $n$-th columns to determine $a_{2}$ and $a_{n^{2}-1}$. Remove these two numbers from the 1st and $n$-th columns and replace them with adjacent blue numbers.
Similarly, determine $a_{3}, a_{4}, \cdots, a_{n^{2}-2}$ using the following two methods:
Operation $A$: When the number of blue numbers in the 1st and $n$-th columns is less than $\frac{n}{2}$, input these blue numbers along with several red (yellow) numbers from the left (right) of the password to form $n$ numbers (take $\frac{n}{2}$ numbers from each of the 1st and $n$-th columns). This will determine the leftmost and rightmost remaining numbers in the password. Remove these two numbers and replace them with adjacent blue numbers.
Operation $B$: When the number of blue numbers in the 1st (or $n$-th) column is $\frac{n}{2}$, input all the numbers in the 1st (or $n$-th) column (re-color them as red (or yellow)) to determine the leftmost (or rightmost) remaining number in the password. Remove this number and replace it with an adjacent blue number.
The $(n+3)$-th operation can be considered as Operation $A$; the $(n+1)$-th and $(n+2)$-th operations can be considered as Operation $B$.
The feasibility of the operations is obvious.
Assume that Operations $A$ and $B$ are performed $x$ and $y$ times, respectively, to determine all $\{a_{i}\}$.
Since each Operation $A$ determines two numbers, each Operation $B$ determines one number, and the last operation determines $n$ or $n-1$ numbers, we have:
$$
2(x-1) + y + n - 1 \leqslant n^{2}.
$$
Consider the number of blue numbers in the 1st and $n$-th columns, which can increase by at most 2 in each Operation $A$ and decrease by $\frac{n}{2}-1$ in each Operation $B$.
If the last operation is Operation $B$, then
$$
2 + 2x \geqslant \left(\frac{n}{2} - 1\right)(y - 2) + 1;
$$
If the last operation is Operation $A$, then
$$
2 + 2(x - 1) \geqslant \left(\frac{n}{2} - 1\right)(y - 2).
$$
From equations (1) and (2), we have
$$
\begin{array}{l}
2x + y - 1 \geqslant \frac{n}{2}(y - 2) \\
\Rightarrow n^{2} - n + 2 \geqslant \frac{n}{2}(y - 2) \\
\Rightarrow \frac{y}{2} \leqslant n + \frac{2}{n}.
\end{array}
$$
Thus, the total number of operations is
$$
\begin{array}{l}
n + x + y = n + \frac{(2x + y) + y}{2} \\
\leqslant n + \frac{n^{2} - n + 3}{2} + n + \frac{2}{n} \\
= \frac{(n + 1)(n + 2) + 1}{2} + \frac{2}{n} = 45 \frac{3}{4} \\
\Rightarrow n + x + y \leqslant 45.
\end{array}
$$
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$
If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers.
|
1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.
Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.
Below, we prove by contradiction: For any odd prime $a$, at least one of $y_{1}, y_{2}, y_{3}$ is composite.
Otherwise, $x_{1}, x_{2}, x_{3}$ are all primes.
Since $x_{1} \geqslant 3$ is odd, we know
$x_{2}>3$, and $x_{2}=3(\bmod 4)$.
Therefore, $x_{3} \equiv 7(\bmod 8)$.
Thus, 2 is a quadratic residue of $x_{3}$, i.e., there exists $x \in \mathbf{N}_{+}$ such that
$$
x^{2}=2\left(\bmod x_{3}\right) \text {. }
$$
Therefore, $2^{x_{2}}=2^{\frac{x_{3}-1}{2}} \equiv x^{x_{3}-1} \equiv 1\left(\bmod x_{3}^{3}\right)$.
Thus, $x_{3} \mid y_{2}$.
Since $x_{2}>3$, then $2^{x_{2}}-1>2 x_{2}+1=x_{3}$.
Therefore, $y_{2}$ is composite.
Finally, if $a=2$, then
$$
y_{1}=3, y_{2}=31 \text {, and } 23 \mid y_{3}=\left(2^{11}-1\right) \text {. }
$$
Therefore, $k=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the set
$$
A=\{x \mid 5 x-a \leqslant 0, a \in \mathbf{N}\} \text {. }
$$
If $5 \in A \cap \mathbf{Z}$, then the minimum value of $a$ is
|
$$
-, 1.25
$$
From $A \left\lvert\,=\left(-\infty, \frac{a}{5}\right]\right.$, and $5 \in A \cap \mathbf{Z}$, we know
$$
\frac{a}{5} \geqslant 5 \Rightarrow a \geqslant 25 \text {. }
$$
Therefore, the minimum value of $a$ is 25.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
\begin{array}{l}
f(x+1)=f(-x), \\
f(x)=\left\{\begin{array}{ll}
1, & -1<x \leqslant 0 \\
-1, & 0<x \leqslant 1 .
\end{array}\right.
\end{array}
$$
Then $f(f(3.5))=$ $\qquad$
|
2. -1 .
From $f(x+1)=-f(x)$, we know $f(x+2)=f(x)$.
Then $f(3.5)=f(-0.5)=1$.
Therefore, $f(f(3.5))=f(1)=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
the graph always passes through a fixed point $A$. If point $A$ lies on the line
$$
\frac{x}{m}+\frac{y}{n}+1=0(m, n>0)
$$
then the minimum value of $3 m+n$ is
|
5. 16 .
Note that the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
always passes through the fixed point $(-3,-1)$.
So point $A(-3,-1)$.
Then $-\frac{3}{m}-\frac{1}{n}+1=0 \Rightarrow 1=\frac{3}{m}+\frac{1}{n}$.
Thus, $3 m+n=(3 m+n)\left(\frac{3}{m}+\frac{1}{n}\right)$
$$
=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 \text {. }
$$
When and only when $m=n=4$, the equality holds. Therefore, the minimum value of $3 m+n$ is 16.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $P$ is a moving point on the line $l$:
$$
k x+y+4=0(k>0)
$$
$P A$ and $P B$ are the two tangents from $P$ to the circle $C$:
$$
x^{2}+y^{2}-2 y=0
$$
with points of tangency $A$ and $B$ respectively. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$
|
6.2.
$$
\begin{array}{l}
\text { Given } S_{\text {quadrilateral } P A C B}=P A \cdot A C=P A \\
=\sqrt{C P^{2}-C A^{2}}=\sqrt{C P^{2}-1},
\end{array}
$$
we know that the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$. Also, $\sqrt{C P^{2}-1}=2$, then $C P=\sqrt{5}$. Using the point-to-line distance formula, we get
$$
C P=\sqrt{5}=\frac{5}{\sqrt{1+k^{2}}} \text {. }
$$
Since $k>0$, we have $k=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the shortest distance from $P$ along the side surface of the prism through the edge $C C_{1}$ to point $M$ is $3 \sqrt{2}$. Let this shortest distance intersect $C C_{1}$ at point $N$.
(1) Prove: $A_{1} B / /$ plane $M N P$;
(2) Find the tangent value of the dihedral angle (acute angle) formed by plane $M N P$ and plane $A B C$.
|
10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles.
As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position of $P_{1}$, and connect $M P_{1}$. Then $M P_{1}$ is the shortest path from point $P$ along the lateral surface of the prism through $C C_{1}$ to point $M$.
Let $P C=x$. Then $P_{1} C=x$.
In the right triangle $\triangle M A P_{1}$, note that,
$$
\begin{array}{l}
(2+x)^{2}+3^{2}=18 \\
\Rightarrow x=1 .
\end{array}
$$
Thus, $P$ is the midpoint of $B C$.
Therefore, $N C=1$.
Let $A_{1} C$ intersect $M N$ at point $Q$. Then $Q$ is the midpoint of $A_{1} C$.
So, $A_{1} B / / P Q$.
Hence, $A_{1} B / /$ plane $M N P$.
(2) As shown in Figure 3, connect $C H$ and $P P_{1}$. Then $P P_{1}$ is the intersection line of plane $M N P$ and plane $A B C$.
Draw $N H \perp P P_{1}$ at point $H$.
Since $C C_{1} \perp$ plane $A B C$, it follows that $C H \perp P P_{1}$.
Therefore, $\angle N H C$ is the plane angle of the dihedral angle formed by plane $N M P$ and plane $A B C$.
In the right triangle $\triangle P H C$, we have
$$
\angle P C H=\frac{1}{2} \angle P C P_{1}=60^{\circ} \text {, }
$$
Then $C H=\frac{1}{2}$.
In the right triangle $\triangle N C H$,
$$
\tan \angle N H C=\frac{N C}{C H}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$
and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis, respectively.
(1) Prove: $F<0$;
(2) If the area of quadrilateral $A B C D$ is 8, the length of diagonal $A C$ is 2, and $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, find the value of $D^{2}+E^{2}-4 F$.
|
13. (1) Let $A(a, 0), C(c, 0)$.
From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$
where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1} x_{c}=a c=F<0$.
(2) From the problem, we know
$S_{\text {quadrilateral } A B C D}=\frac{|\overrightarrow{A C}||\overrightarrow{B D}|}{2}=8$.
Also, $|\overrightarrow{A C}|=2$, so $|\overrightarrow{B D}|=8$.
Since $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, we have $\angle A=90^{\circ}$.
Thus, $|\overrightarrow{B D}|=2 r=8 \Rightarrow r=4$.
From the circle represented by equation (1), we have
$$
\frac{D^{2}}{4}+\frac{E^{2}}{4}-F=r^{2} \text{. }
$$
Therefore, $D^{2}+E^{2}-4 F=4 r^{2}=64$.
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
I. Fill in the Blanks (8 points each, total 64 points)
1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$.
|
$-, 1.327$
$$
C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327
$$
|
327
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$
|
3. 1 .
From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\triangle A B C$.
Hence, $M O=\sqrt{M C^{2}-O C^{2}}=1$ is the desired result.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$
|
6. 54.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$
and $q>1$.
Let $t=q^{2}-1$. Then
$$
\begin{array}{l}
2 a_{8}+a_{7}=2 a_{2} q^{6}+a_{1} q^{6} \\
=\left(2 a_{2}+a_{1}\right) q^{6} \\
=\frac{8 q^{6}}{q^{2}-1}=\frac{8(t+1)^{3}}{t} \\
=8\left(t^{2}+3 t+3+\frac{1}{t}\right) .
\end{array}
$$
Let $f(t)=8\left(t^{2}+3 t+3+\frac{1}{t}\right)$.
Then $f^{\prime}(t)=8\left(2 t+3-\frac{1}{t^{2}}\right)$
$$
=8 \cdot \frac{(t+1)^{2}(2 t-1)}{t^{2}} \text {. }
$$
Thus, $f(t)$ is a decreasing function on $\left(0, \frac{1}{2}\right)$ and an increasing function on $\left(\frac{1}{2},+\infty\right)$.
Therefore, $f(t)_{\min }=f\left(\frac{1}{2}\right)=54$.
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$
|
8. 18084.
When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 1023\right] \\
=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9} .
\end{array}
$$
Let $S=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9}$.
Then $2 S=1 \times 2^{2}+2 \times 2^{3}+\cdots+9 \times 2^{10}$.
Subtracting the above two equations, we get
$$
\begin{array}{l}
S=-2-2^{2}-\cdots-2^{9}+9 \times 2^{10} \\
=-2^{10}+2+9 \times 2^{10}=8194
\end{array}
$$
Therefore, the original expression $=8194+9890=18084$.
|
18084
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $[x]$ denote the integer part of the real number $x$. Then $[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]=(\quad)$.
(A) 146
(B) 161
(C) 210
(D) 365
|
4. C.
$$
\begin{array}{l}
{[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]} \\
=1 \times 3+2 \times 5+3 \times 7+4 \times 9+5 \times 11+6 \times 13+7 \\
=210
\end{array}
$$
|
210
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 From the 205 positive integers $1,2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a<b<c)$, we have
$$
a b \neq c ?^{[1]}
$$
(2005, (Casio Cup) National Junior High School Mathematics Competition)
|
Estimate first.
Since $14 \times 15=210>205$, then $14,15, \cdots$, 205 satisfy that for any three numbers $a 、 b 、 c(a<b<c)$, we have $a b \neq c$.
Because 1 multiplied by any number equals the number itself, $1,14,15, \cdots, 205$ satisfy the condition.
Therefore, there are $205-14+1+1=193$ numbers in total.
If we select one more number from $2 \sim 13$, what would happen?
Construct the following 12 sets:
$(2,25,2 \times 25),(3,24,3 \times 24), \cdots$,
$(13,14,13 \times 14)$.
The 36 numbers in the above sets are all distinct, and the smallest number is 2, while the largest number is $13 \times 14=182<205$.
Therefore, all three numbers in each set cannot be selected.
In summary, the number of numbers that satisfy the given condition does not exceed $205-12=193$.
|
193
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, ..., and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest scores of each athlete's nine scores is no more than 3. Let the total scores of each athlete be $c_{1}, c_{2}, \cdots, c_{12}$, and $c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{12}$. Find the maximum value of $c_{1}$.
(2004, Jiangsu Province Junior High School Mathematics Competition (Grade 9))
|
Explanation: It is impossible for 9 judges to give 1 point to five or more athletes, because among five or more athletes, at least one athlete must be rated no less than 5 by a judge. However, according to the problem, each of these five athletes is rated no more than 4 by each judge, which is a contradiction.
Therefore, 9 judges can give 1 point to at most four athletes.
We will discuss the following scenarios.
(1) If all judges give 1 point to a single athlete, then \( c_{1} = 9 \).
(2) If the nine 1 points given by the 9 judges are concentrated on two athletes, then one of these athletes must be rated 1 by at least five judges. According to the problem, the remaining judges give this athlete a score no greater than 4, thus,
\[
c_{1} \leqslant 5 \times 1 + 4 \times 4 = 21.
\]
(3) If the nine 1 points are concentrated on three athletes, then the total score of these three athletes is no more than
\[
9 \times 1 + 9 \times 3 + 9 \times 4 = 72.
\]
Thus, \( 3 c_{1} \leqslant c_{1} + c_{2} + c_{3} \leqslant 72 \).
Hence, \( c_{1} \leqslant 24 \).
(4) If the nine 1 points are distributed among four athletes, then the total score of these four athletes is
\[
9 \times 1 + 9 \times 2 + 9 \times 3 + 9 \times 4 = 90.
\]
Thus, \( 4 c_{1} \leqslant 90 \). Hence, \( c_{1} < 23 \).
In summary, \( c_{1} \leqslant 24 \).
The scenario where \( c_{1} = 24 \) is achievable. Let \( A_{i} (i = 1, 2, \cdots, 12) \) represent the athletes, and \( B_{j} (j = 1, 2, \cdots, 9) \) represent the judges. The values in Table 1 are the scores given by judge \( B_{j} \) to athlete \( A_{i} \).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & \( A_{1} \) & \( A_{2} \) & \( A_{3} \) & \( A_{4} \) & \( A_{5} \) & \( A_{6} \) & \( A_{7} \) & \( A_{8} \) & \( A_{9} \) & \( A_{10} \) & \( A_{11} \) & \( A_{12} \) \\
\hline \( B_{1} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{2} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{3} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{4} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{5} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{6} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{7} \) & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline \( B_{8} \) & 3 & 1 & 4 & 5 & 2 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline \( B_{9} \) & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline Total & 24 & 24 & 24 & 30 & 33 & 66 & 66 & 66 & 87 & 87 & 87 & 108 \\
\hline
\end{tabular}
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$.
|
(1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN \text{. }
$$
Thus, quadrilaterals $ABCM$ and $ABNC$ are both isosceles trapezoids, and we have
$$
BM=AC=b, AN=BC=a.
$$
Note that $M$ and $N$ are the midpoints of arcs $\overparen{AC}$ and $\overparen{BC}$ (excluding $B$ or $A$), respectively. By Corollary 2(2), we have
$$
\begin{array}{l}
MB^2 - MA^2 = AB \cdot BC, \\
NA^2 - NB^2 = AB \cdot AC,
\end{array}
$$
which gives $b^2 - c^2 = ac, a^2 - b^2 = bc$.
Adding these two equations, we get
$$
\begin{array}{l}
a^2 - ac - c(b+c) \\
\Rightarrow \frac{b+c}{a} = \frac{a-c}{c} = \frac{a}{c} - 1.
\end{array}
$$
Also, from $b^2 + bc = a^2$, we have
$$
\frac{b+c}{a} = \frac{a}{b}.
$$
From equations (1) and (2), we get
$$
\frac{a}{c} - \frac{a}{b} = 1 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{c}.
$$
(2) Solution From equation (3), we have $ab = ac + bc$.
Thus, $\frac{(a+b-c)^2}{a^2 + b^2 + c^2}$
$$
= \frac{a^2 + b^2 + c^2 + 2(ab - bc - ac)}{a^2 + b^2 + c^2} = 1.
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the equation in $x$
$$
(a-1) x^{2}+2 x-a-1=0
$$
has roots that are all integers. Then the number of integer values of $a$ that satisfy this condition is
$\qquad$.
|
4.5.
When $a=1$, $x=1$.
When $a \neq 1$, it is easy to see that $x=1$ is an integer root of the equation.
Furthermore, from $1+x=\frac{2}{1-a}$ and $x$ being an integer, we know
$$
1-a= \pm 1, \pm 2 \text {. }
$$
Therefore, $a=-1,0,2,3$.
In summary, there are 5 integer values of $a$ that satisfy the condition.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$
holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, then $f(x+t) \leqslant x$ holds.
|
(1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$
Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4}(x+1)^{2}$.
(2) Suppose there exists $t \in \mathbf{R}$, such that for $1 \leqslant x \leqslant m$, we have $f(x+t) \leqslant x$.
Taking $x=1$, we have
$$
(t+2)^{2} \leqslant 4 \Rightarrow-4 \leqslant t \leqslant 0 \text {. }
$$
Taking $x=m$, we have
$$
\begin{array}{l}
(m+t+1)^{2} \leqslant 4 m \\
\Rightarrow m^{2}+2(t-1) m+\left(t^{2}+2 t+1\right) \leqslant 0 \\
\Rightarrow 1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t} .
\end{array}
$$
Thus, $m \leqslant 1-t+\sqrt{-4 t}$
$$
\leqslant 1-(-4)+\sqrt{-4 \times(-4)}=9 \text {. }
$$
When $t=-4$, for any $x(1 \leqslant x \leqslant 9)$, we always have
$$
\begin{array}{l}
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right) . \\
=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
\end{array}
$$
Therefore, the maximum value of $m$ is 9.
(Sun Yan, Anhui Provincial Anqing City Educational Research Office, 246003, Huang Quanfu, Anhui Provincial Huaining County Jiangzhen Middle School, 246142)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ ـ.
|
7.8.
First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known sequence such that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1 .
$$
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
It is easy to see that for any $n$, $x_{n}=2(\bmod 3)$.
Thus, taking modulo 3 on both sides of equation (1) gives
$$
2^{k-1} k \equiv 2^{k}(\bmod 3) \Rightarrow k \equiv 2(\bmod 3) \text {. }
$$
When $k=2$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1 \text {; }
$$
When $k=5$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1 \text {; }
$$
When $k=8$, taking $2, 5, 8, 11, 20, 41, 110, 1640$ satisfies the condition.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Given the sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right],
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
$$
\text { Find } \sum_{i=1}^{2012} a_{i} \text {. }
$$
|
$$
\text { II. } a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}, a_{4}=\frac{5}{3} .
$$
By mathematical induction, it is easy to prove
$$
\left\{\begin{array}{l}
a_{2 k+1}=\frac{3 k+1}{3}, \\
a_{2 k+2}=\frac{3 k+2}{3} .
\end{array}\right.
$$
Then $a_{2 k+1}+a_{2 k+2}=2 k+1$.
Therefore, $\sum_{i=1}^{2012} a_{i}=1012036$.
|
1012036
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Can 2010 be written as the sum of squares of $k$ distinct positive integers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. ${ }^{[2]}$
(2010, Beijing Middle School Mathematics Competition (Grade 8))
|
Estimate the approximate range of $k$ first, then discuss by classification.
Let $p_{i}$ be a prime number.
If 2010 can be written as the sum of squares of $k$ prime numbers, then by
$$
\begin{array}{l}
2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}+19^{2}+23^{2}+29^{2} \\
=2397>2010,
\end{array}
$$
we know $k \leqslant 9$.
(1) When $k=9$, if
$$
2010=p_{1}^{2}+p_{2}^{2}+\cdots+p_{9}^{2} \text {, }
$$
among them, there must be one square of an even number and eight squares of odd prime numbers.
The left side of the equation is congruent to 2 modulo 8, while the right side is congruent to 4 modulo 8, so the equation cannot hold.
(2) When $k=8$, if
$$
2010=p_{1}^{2}+p_{2}^{2} \cdots+p_{8}^{2} \text {, }
$$
these eight addends are all squares of odd prime numbers.
The left side of the equation is congruent to 2 modulo 8, while the right side is congruent to 0 modulo 8, so the equation cannot hold.
(3) When $k=7$, after trial calculation, we find
$$
2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}=2010 \text {. }
$$
In summary, the maximum value of $k$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Choose $k$ numbers from $1,2, \cdots, 2004$, such that among the chosen $k$ numbers, there are definitely three numbers that can form the side lengths of a triangle (the three numbers must be distinct). What is the minimum value of $k$ that satisfies this condition?
|
When selecting three numbers from 1 to 2004 to form the sides of a triangle, there are too many possibilities. Instead, let's approach it from the opposite direction and list all sets of three numbers that cannot form the sides of a triangle.
First, 1, 2, 3 cannot form the sides of a triangle. Adding 5, the set \(1, 2, 3, 5\) also cannot form the sides of a triangle.
Continuing this way, the added number is exactly the sum of the last two numbers, resulting in:
\[
\begin{array}{l}
1, 2, 3, 5, 8, 13, 21 ; 34, 55, 89, 144, 233, \\
377, 610, 987, 1597
\end{array}
\]
There are 16 numbers in total. In these 16 numbers, any three chosen cannot form the sides of a triangle. If we add any other number from the remaining ones, then among these 17 numbers, we can find three numbers that can form the sides of a triangle.
Therefore, the smallest value of \(k\) that satisfies the condition is 17.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A. Let $a=\sqrt[3]{3}, b$ be the fractional part of $a^{2}$. Then $(b+2)^{3}=$ $\qquad$
|
ニ、6. A.9.
From $2<a^{2}<3$, we know $b=a^{2}-2=\sqrt[3]{9}-2$. Therefore, $(b+2)^{3}=(\sqrt[3]{9})^{3}=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$
Then the maximum value of $a b c$ is $\qquad$
|
8. A. 2013.
The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$
Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, which has no positive integer solutions; if $a=3$, then $(b-8)^{2}=9$. Solving this, we get $b=11$ or 5. (1) If $b=11$, then $c=61$, thus, $a b c=3 \times 11 \times 61=2013$;
(2) If $b=5$, then $c=13$, thus, $a b c=3 \times 5 \times 13=195$.
In summary, the maximum value of $a b c$ is 2013.
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. If $k$ numbers are chosen from 2, $, 8, \cdots, 101$ these 34 numbers, where the sum of at least two of them is 43, then the minimum value of $k$ is: $\qquad$
|
B. 28.
Divide the 14 numbers less than 43 into the following seven groups: $(2,41),(5,38),(8,35),(11,32)$, $(14,29),(17,26),(20,23)$.
The sum of the two numbers in each group is 43. After selecting one number from each group and then taking all numbers greater than 43, a total of 27 numbers are selected. The sum of any two of these 27 numbers is not equal to 43.
On the other hand, if 28 numbers are selected, then in the above groups, there must be one group where both numbers are selected, and their sum is 43.
In summary, the minimum value of $k$ is 28.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A. Xiaoming volunteered to sell pens at a stationery store one day. Pencils were sold at 4 yuan each, and ballpoint pens at 7 yuan each. At the beginning, it was known that he had a total of 350 pencils and ballpoint pens. Although he did not sell them all that day, his sales revenue was 2013 yuan. Then he sold at least $\qquad$ ballpoint pens.
|
10. A. 207.
Let $x$ and $y$ represent the number of pencils and ballpoint pens sold, respectively. Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
4 x+7 y=2013 ; \\
x+y=204
\end{array}\right.
\end{array}
\]
Thus, $y_{\text {min }}=207$, at which point, $x=141$.
|
207
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} 、 y_{0}$ are integers, and points $O 、 A 、 B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$.
|
B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}-\frac{y_{0}}{10}\right| \\
=5\left|10 x_{0}-y_{0}\right| \neq 0 .
\end{array}
$$
Given that $\left|10 x_{0}-y_{0}\right|$ is an integer, then $\left|10 x_{0}-y_{0}\right| \geqslant 1$.
Therefore, $S_{\triangle O A B}=5\left|10 x_{0}-y_{0}\right| \geqslant 5$.
When $x_{0}=1, y_{0}=9$, $S_{\triangle O A B}=5$.
Thus, the minimum area of $\triangle O A B$ is 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. A. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $C$ is a point on the circumference, and $D$ is a point on the line segment $O B$ (not at the endpoints), satisfying $C D \perp A B$ and $D E \perp C O$ at point $E$. If $C E = 10$, and the lengths of $A D$ and $D B$ are both positive integers, find the length of line segment $A D$.
|
12. A. Connect $A C$ and $B C$, then $\angle A C B=90^{\circ}$.
From $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C O D$, we know $C E \cdot C O=C D^{2}$.
From $\mathrm{Rt} \triangle A C D \backsim \mathrm{Rt} \triangle C B D$, we know $C D^{2}=A D \cdot B D$.
Therefore, $C E \cdot C O=A D \cdot B D$.
Let $A D=a, D B=b\left(a, b \in \mathbf{N}_{+}\right)$. Then $C O=\frac{a+b}{2}$.
Given $C E=10$, substituting into equation (1) yields
$$
10 \times \frac{a+b}{2}=a b \Rightarrow(a-5)(b-5)=25 \text {. }
$$
Considering $a>b$, we can only have $a-5>b-5>0$, leading to $a-5=25, b-5=1$.
Thus, $A D=a=30$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A. If placing the positive integer $M$ to the left of the positive integer $m$ results in a new number that is divisible by 7, then $M$ is called the "magic number" of $m$ (for example, placing 86 to the left of 415 results in the number 86415, which is divisible by 7, so 86 is called the magic number of 415). Find the smallest positive integer $n$ such that there exist distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying that for any positive integer $m$, at least one of $a_{1}, a_{2}, \cdots, a_{n}$ is a magic number of $m$.
|
14. A. If $n \leqslant 6$, take $m=1,2, \cdots, 7$. By the pigeonhole principle, there must be a positive integer $M$ among $a_{1}, a_{2}, \cdots, a_{n}$ that is a common magic number of $i$ and $j (1 \leqslant i<j \leqslant 7)$, i.e.,
$$
7 \mid(10 M+i), 7 \mid(10 M+j) \text {. }
$$
Then $7 \mid (j-i)$. But $0<j-i \leqslant 6$, which is a contradiction. Therefore, $n \geqslant 7$.
Furthermore, when $a_{1}, a_{2}, \cdots, a_{n}$ are $1,2, \cdots, 7$, for any positive integer $m$, let it be a $k$-digit number ($k$ is a positive integer).
Then the remainders of $10^{k} i+m (i=1,2, \cdots, 7)$ when divided by 7 are all different. If not, there exist positive integers $i, j (1 \leqslant i<j \leqslant 7)$, such that
$$
\begin{array}{l}
7 \mid\left[\left(10^{k} j+m\right)-\left(10^{k} i+m\right)\right] \\
\Rightarrow 7 \mid 10^{k}(j-i) \Rightarrow 7 \mid (j-i),
\end{array}
$$
which is a contradiction.
Therefore, there must exist a positive integer $i (1 \leqslant i \leqslant 7)$ such that $7 \mid \left(10^{k} i+m\right)$, i.e., $i$ is a magic number of $m$. Thus, the minimum value of $n$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ .
|
2. -1043 .
From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$
Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$
|
-1043
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+9+\sqrt{x-7}}+\frac{|x+y-z|}{4}=4 \text {, }
$$
then the units digit of $(5 x+3 y-3 z)^{2013}$ is $\qquad$
|
3. 4 .
It is known that $x \geqslant 7$, then
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}} \geqslant 4, \\
\frac{|x+y-z|}{4} \geqslant 0 .
\end{array}\right.
$$
Combining the given equations, we have
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}}=4, \\
\frac{|x+y-z|}{4}=0 .
\end{array}\right.
$$
Therefore, $x=7, x+y-z=0$.
Thus, $(5 x+3 y-3 z)^{2013}=14^{2013}$, and its unit digit is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 3, square $ABCD$ is divided into 8 triangles of equal area. If $AG=\sqrt{50}$, then the area $S$ of square $ABCD$ is $ \qquad $.
|
4. 128 .
As shown in Figure 5, draw $K L / / D C$ through point $F$, take the midpoint $N$ of $A B$, and connect $G N$ with $A H$ intersecting at point $P$.
Let the side length of the square $A B C D$ be $a$.
Given $S_{\triangle D C I}=S_{\triangle M B H}=\frac{1}{8} S$, we know
$C I=B H=\frac{1}{4} B C=\frac{a}{4}$.
Since $S_{\triangle M D F}=2 S_{\triangle D C I}$, we have
$A D \cdot K F=2 C D \cdot C I$.
Therefore, $K F=2 C I=\frac{1}{2} a$.
Thus, $F$ is the midpoint of $D I$.
Given $S_{\triangle C A H}=S_{\triangle C H F}=S_{\triangle C F A}$, we know $G$ is the centroid of $\triangle F A H$.
Then $E$ and $P$ are the midpoints of $A F$ and $A H$ respectively, and
$$
G P=\frac{1}{2} F G \text {. }
$$
Also, $F P$ is the midline of trapezoid $A H I D$, so,
$F P=\frac{H I+A D}{2}=\frac{\frac{1}{2} a+a}{2}=\frac{3 a}{4}$
$\Rightarrow G P=\frac{1}{3} F P=\frac{a}{4}$
$\Rightarrow G N=G P+P N=\frac{a}{4}+\frac{a}{8}=\frac{3 a}{8}$.
Since $A N=\frac{a}{2}$, by the Pythagorean theorem we get
$$
\begin{array}{l}
A G^{2}=\left(\frac{a^{2}}{2}\right)+\left(\frac{3 a}{8}\right)^{2}=\frac{25 a^{2}}{64}=50 \\
\Rightarrow a^{2}=128 \Rightarrow S=128 .
\end{array}
$$
|
128
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$
|
5.11 .
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$
Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rightarrow a=13, b=2 \\
\Rightarrow a-b=11
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$
|
3. 2 .
Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
According to the problem, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O C}=\overrightarrow{O B} \cdot \overrightarrow{O C} \text {, }
$$
which means $3 a+4=6+4 b$.
Thus, $3 a-4 b=2$.
Solution 2 Since the projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, $A B \perp O C$, that is,
$$
\overrightarrow{A B} \cdot \overrightarrow{O C}=0 \text {. }
$$
Therefore, $3(2-a)+4(b-1)=0$, which simplifies to $3 a-4 b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$
|
5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$
Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=3 y<44 \\
\Rightarrow \frac{40}{3}<y<\frac{44}{3} \Rightarrow y=14 \\
\Rightarrow d=-6 y=-84 \\
\Rightarrow q=\frac{y+z}{x+y}=\frac{2 y+d}{2 y-d}=-\frac{1}{2} \\
\Rightarrow d p=42 .
\end{array}
$$
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High School Mathematics Competition
|
Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\frac{a_{j}-a_{i}}{n-1} \text {. }
$$
Therefore, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, we have
$$
(n-1) \mid\left(2^{3} \times 251\right) \text {. }
$$
Also, $a_{n}-1=\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)$
$$
\geqslant \sum_{i=1}^{n-1}(n-1)=(n-1)^{2} \text {, }
$$
so $(n-1)^{2} \leqslant 2008$, which implies $n \leqslant 45$.
Combining with equation (1), we get $n \leqslant 9$.
On the other hand, let
$$
a_{i+1}=8 i+1(i=0,1, \cdots, 7), a_{9}=8 \times 251+1 \text {. }
$$
Then these nine numbers satisfy the given conditions.
Hence, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If for any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$$
\frac{x}{1+x-2 x^{2}}=\sum_{k=0}^{\infty} a_{k} x^{k},
$$
then $a_{3}+a_{4}=$ . $\qquad$
|
9. -2 .
In equation (1), let $x=0$, we get $a_{0}=0$. Then $\frac{1}{1+x-2 x^{2}}=\sum_{k=1}^{\infty} a_{k} x^{k-1}$. Substituting $x=0$ into the above equation, we get $a_{1}=1$. Then $\frac{-1+2 x}{1+x-2 x^{2}}=\sum_{k=2}^{\infty} a_{k} x^{k-2}$. Substituting $x=0$ into the above equation, we get $a_{2}=-1$. Similarly, $a_{3}=3, a_{4}=-5$. Therefore, $a_{3}+a_{4}=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$, then
$$
M=x_{1}-x_{2}^{3}+x_{2}-x_{3}^{3}+x_{3}-x_{4}^{3}+x_{4}-x_{5}^{3}+x_{5}-x_{1}^{3}
$$
the maximum value is $\qquad$
|
10.4.
If there exists a positive integer $j$, such that
$$
x_{j}=x_{j+1}\left(j=1,2, \cdots, 5, x_{6}=x_{1}\right) \text {, }
$$
then $M \leqslant 4$.
When $x_{1}=0, x_{2}=1, x_{3}=0, x_{4}=1, x_{5}=0$, the equality holds.
If for any positive integer $i$, we have $x_{i} \neq x_{i+1}$ $(i=1,2, \cdots 5)$, then either $x_{i-1}x_{i+1}$, or $x_{i-1}>x_{i}$ and $x_{i}<x_{i+1}$.
For both cases, we have
$$
\left|x_{i-1}-x_{i}\right|^{3}+\left|x_{i}-x_{i+1}\right|^{3}<\left|x_{i-1}-x_{i+1}\right|^{3} \leqslant 1 \text {. }
$$
Thus, $M<1+3=4$.
In summary, the maximum value of $M$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are three sets of cards in red, yellow, and blue, each set containing five cards, marked with the letters $A, B, C, D, E$. If five cards are drawn from these 15 cards, with the requirement that the letters are all different and all three colors are included, then the number of different ways to draw the cards is $\qquad$ kinds.
|
4. 150.
Divide into two categories: $3, 1,1$ and $2,2,1$.
Calculate respectively:
$$
\frac{C_{3}^{1} C_{5}^{3} C_{2}^{1} C_{2}^{1} C_{1}^{1}}{A_{2}^{2}}+\frac{C_{3}^{1} C_{5}^{2} C_{2}^{1} C_{3}^{2} C_{1}^{1}}{A_{2}^{2}}=150
$$
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$
Then the area of the figure formed by the moving point $P(x, y)$ is
|
8. 2 .
From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$
It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{array}{l}
\text { Therefore, equation (1) } \Leftrightarrow f(x) \geqslant f(-y) \\
\Leftrightarrow x \geqslant -y \Leftrightarrow x+y \geqslant 0 .
\end{array}
$$
From $\left\{\begin{array}{l}2 x+y \leqslant 2 \\ x+y \geqslant 0\end{array}\right.$, we get the intersection point coordinates as $(2,-2)$.
Thus, the area of the region formed by the moving point $P$ is
$$
\frac{1}{2} \times 2 \times 2=2
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=2012 ?
$$
|
14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.
Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$
Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is an integer greater than 1.
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Clearly, $(n, 3 n+1)=1$.
Hence, $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Also, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$, which is a contradiction.
Therefore, $f(n) \neq 2$.
When $n=8$,
$$
3 n^{2}+n+1=201 \text {. }
$$
Thus, $f(8)=3$.
In summary, the minimum value of $f(n)$ is 3.
(2) When $n=2 \times 10^{k}-1$, we have
$$
\begin{array}{l}
3 n^{2}+n+1=12 \times 10^{2 k}-10 \times 10^{k}+3 \\
=11 \underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 03}_{k \uparrow} .
\end{array}
$$
Thus, $f(n)=1+1+9(k-1)+3$
$$
=9 k-4\left(k \in \mathbf{N}_{+}\right) \text {. }
$$
(3) There exists a positive integer $n$ such that
$$
f(n)=2012 \text {. }
$$
In fact, let $n=2 \times 10^{k}-1$. Then $f(n)=9 k-4$.
Let $9 k-4=2012$. Solving for $k$ gives $k=224$.
Therefore, taking $n=2 \times 10^{224}-1$, we get
$$
f(n)=2012 \text {. }
$$
(Zhang Shengchun provided)
|
2012
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The positive integer $n=$ $\qquad$ that makes $2^{n}+256$ a perfect square.
|
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$.
If it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a positive integer). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, hence $k=1$.
Thus, $n=11$.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-digit or six-digit number in $B$;
(3) If the elements in $B$ are arranged in ascending order, find the 1081st element.
|
(1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers that meet the criteria is $9 \times 8=72$.
For three-digit numbers, first consider the digit in the hundreds place, which can be $1,2, \cdots, 9$. Next, consider the digit in the tens place, which cannot be the same as the digit in the hundreds place and the sum of the two digits cannot be 9, so there are 8 possible choices. Finally, consider the digit in the units place, which cannot be the same as the digits in the hundreds and tens places, and the sum with the digits in the hundreds and tens places cannot be 9, so there are 6 possible choices.
Therefore, the number of three-digit numbers that meet the criteria is
$$
9 \times 8 \times 6=432 \text{ (numbers). }
$$
(2) There exists a five-digit number that meets the criteria (e.g., 12340).
There does not exist a six-digit number that meets the criteria. The reason is as follows.
Following the method in (1), there are 9 possible choices for the hundred-thousands place, 8 for the ten-thousands place, 6 for the thousands place, 4 for the hundreds place, 2 for the tens place, and 0 for the units place, which is a contradiction.
(3) From (1), we know that there are 72 two-digit numbers and 432 three-digit numbers that meet the criteria.
There are also
$$
9 \times 8 \times 6 \times 4=1728
$$
numbers that meet the criteria among four-digit numbers, and among these, there are $8 \times 6 \times 4=192$ numbers with 1 in the thousands place, and 192 numbers each with 2 or 3 in the thousands place.
Since $1081-(72+432+192 \times 3)=1$, the number that meets the criteria is the smallest number with 4 in the thousands place, i.e., 4012 is the 1081st element of $B$.
|
4012
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 For a four-digit number, at most two of its digits are different. How many such four-digit numbers are there?
保留源文本的换行和格式，所以翻译结果如下：
Example 2 For a four-digit number, at most two of its digits are different.
Ask: How many such four-digit numbers are there?
|
Solution: Clearly, there are exactly 9 four-digit numbers where all four digits are the same.
Below, we consider four-digit numbers with exactly two different digits in three steps.
(1) First, consider the thousands place, which has 9 possible choices: $1,2, \cdots, 9$.
(2) Next, consider the hundreds, tens, and units places. Since there are exactly two different digits, choose 1 digit from $0,1, \cdots, 9$ in addition to the thousands place digit.
(3) After determining the two digits in the first two steps, further determine the digits in the hundreds, tens, and units places. Each of these positions has two choices, but one scenario must be excluded, where the digits in the hundreds, tens, and units places are all the same as the thousands place digit, so there are $2 \times 2 \times 2-1$ ways to choose.
In summary, there are a total of four-digit numbers
$$
9+9 \times 9(2 \times 2 \times 2-1)=576 \text { (numbers). }
$$
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $A \cup B \cup C=\{1,2, \cdots, 6\}$, and $A \cap B=\{1,2\},\{1,2,3,4\} \subseteq B \cup C$.
Then the number of $(A, B, C)$ that satisfy the conditions is $\qquad$ groups (different orders of $A, B, C$ are considered different groups).
|
As shown in Figure 1, for $1$ and $2$, they can belong to regions I and II, which gives $2^{2}$ possibilities; for $3$ and $4$, they can belong to $B \cup C$ except for regions I and II, i.e., regions III, IV, VI, and VII, which gives $4^{2}$ possibilities; for $5$ and $6$, they can belong to $A \cup B$ except for regions I and II, i.e., regions III, V, V, VI, and VI, which gives $5^{2}$ possibilities. Therefore, there are a total of $2^{2} \times 4^{2} \times 5^{2}=1600$ combinations.
|
1600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999?
|
(1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$
is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.
If the sum of the digits is 2 and it is an odd number, then the first and last digits must both be 1, with all the middle digits being 0. Let's assume it is $10^{a}+1$ $\left(a \in \mathbf{Z}_{+}\right)$.
$$
\begin{array}{l}
\text { Hence } 3 n^{2}+n+1=10^{a}+1 \\
\Rightarrow n(3 n+1)=10^{a} . \\
\text { But }(3 n+1, n)=(1, n)=1, \text { and } 3 n+1>n,
\end{array}
$$
Thus, $n=2^{a}, 3 n+1=5^{a}$.
So, $3 \times 2^{a}+1=5^{a}$.
But when $a>1$,
$$
\left(\frac{5}{2}\right)^{a}>3+1>3+\frac{1}{2^{a}},
$$
which is a contradiction.
Therefore, $a$ can only be 1.
Thus, $n=2$, and in this case, $3 n+1=7 \neq 5$, which is a contradiction.
Therefore, the sum of the digits must be at least 3.
When $n=8$, $3 n^{2}+n+1=201$, and the sum of the digits is exactly 3, so 3 is the minimum value.
$$
\begin{array}{l}
\text { (2) Let } n=10^{m}-1 \text {. Then } \\
3 n^{2}+n+1=3 \times 10^{2 m}-6 \times 10^{m}+3+10^{m} \\
=2 \underbrace{99 \cdots 95}_{m-1 \uparrow} \underbrace{00 \cdots 03}_{m-1 \uparrow} . \\
\text { Let } 2+9(m-1)+5+3=1999 .
\end{array}
$$
So, we only need to take $m=222$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If 6 college graduates apply to three employers, and each employer hires at least one of them, then the number of different hiring scenarios is $\qquad$ kinds.
|
The number of ways to hire 3 people is $\mathrm{A}_{6}^{3}=120$; the number of ways to hire 4 people is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1} \mathrm{C}_{3}^{1}}{2!} \times \mathrm{A}_{3}^{3}=540$; the number of ways to hire 5 people is
$$
\frac{C_{6}^{2} C_{4}^{2} C_{2}^{1}}{2!} \times A_{3}^{3}+\frac{C_{6}^{3} C_{3}^{1} C_{2}^{1}}{2!} \times A_{3}^{3}=900 \text { (ways); }
$$
The number of ways to hire 6 people is
$$
\begin{array}{l}
\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{2}}{3!} \times \mathrm{A}_{3}^{3}+\frac{\mathrm{C}_{6}^{4} \mathrm{C}_{2}^{1} \mathrm{C}_{1}^{1}}{2!} \times \mathrm{A}_{3}^{3}+\mathrm{C}_{6}^{3} \mathrm{C}_{3}^{2} \mathrm{C}_{1}^{1} \mathrm{~A}_{3}^{3} \\
=540 \text { (ways). }
\end{array}
$$
In summary, there are $120+540+900+540=2100$ ways.
|
2100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 In a $6 \times 6$ grid, three identical red cars and three identical black cars are parked, with one car in each row and each column, and each car occupies one cell. The number of ways to park the cars is ( ).
(A) 720
(B) 20
(C) 518400
(D) 14400
|
Assume first that the three red cars are distinct and the three black cars are also distinct. The first car can obviously be placed in any of the 36 squares, giving 36 ways. The second car, which cannot be in the same row or column as the first car, has 25 ways to be placed.
Similarly, the third, fourth, fifth, and sixth cars have $16$, $9$, $4$, and $1$ ways to be placed, respectively.
Noting that the three red cars are identical and the three black cars are also identical, we have
$$
\frac{36 \times 25 \times 16 \times 9 \times 4 \times 1}{3! \times 3!} = (5!)^2 = 14400
$$
different ways. Therefore, the answer is (D).
|
14400
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
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